Name:___________________________________
Homework #3 (Day)
Due: Mon, Jun 13, 11 am
Read the directions carefully, and answer all parts of each question. Write neatly; if I can’t read it, it will
be counted wrong. You may work in groups, but you must submit your own solution. Any submissions
that are word-for-word identical will receive a score of 0. Use this sheet as a cover page or lose five
points. No late homework will be accepted under any circumstances.
Table for Problem 1
Incoming Freshman
Current Freshman
Current Sophomore
Economics
3
2
4
Business
17
15
13
Accounting
11
20
9
1. Using the table above…
a. Calculate all joint and marginal probabilities.
Table for Problem 1
Economics
Incoming Freshman
0.031914894 0.180851064
Current Freshman
0.021276596 0.159574468
Current Sophomore
0.042553191 0.138297872
b.
c.
d.
e.
f.
g.
Business
Accounting
0.117021277 0.329787
0.212765957 0.393617
0.095744681 0.276596
0.425531915
0.095744681 0.478723404
The probabilities in red are the joints.
The probabilities in green are the marginals by major
The probabilities in blue are the marginals by class.
Calculate the probability that a student is a sophomore in Business.
This is simply the joint: 0.138
Calculate the probability that a student is not a current freshman in Economics.
You can add all joints that are not relevant or simply take the complement.
1 – 0.021 = 0.979
Calculate the probability that a student is a Business major
Simply use the marginal: 0.478
Calculate the probability that a student is either an incoming freshman or in Accounting.
You need the addition rule. Add the two marginals and subtract the joint to remove the
double counting: 0.329 + 0.425 – 0.117 = 0.637
Given that a student is an Economics major, calculate the probability that he/she is a
current sophomore.
This is a conditional. You’ll divide the joint of the two by the marginal of what you know
which is “Economics”. 0.042/0.095 = 0.442
Given that a student is currently enrolled, calculate the probability that he/she is not in
Accounting.
Another conditional, however with a twist. You need the joint of currently enrolled and
not Accounting. This can be achieved a number of ways. What it boils down to is the
sum of four joints: CurFre/Econ, CurFre/Bus, CurSop/Econ, and CurSop/Bus which is
“It’s supposed to be hard. Otherwise, everyone would do it.
The hard is what makes it great.”
-Tom Hanks as Jimmy Dugan, A League of Their Own
Name:___________________________________
Homework #3 (Day)
Due: Mon, Jun 13, 11 am
0.36. Now I need the marginal of currently enrolled which is the sum of the marginals of
CurFre and CurSop or 0.669. Now it’s division: 0.36/0.669 = 0.538. Fun stuff!!
2. Base all of your answers on a standard deck of 52 cards.
a. The probability of drawing a club on a single draw.
13 clubs, 52 cards. 13/52 or 0.25
b. The probability of drawing a six or seven on a single draw.
8 possible successes, 52 cards, 8/52 or 0.153
c. The probability of not drawing a face card on a single draw.
12 face cards, so let’s use the complement. 40 cards are not face cards, so 40/52 or
0.769
d. The probability of drawing a heart or a spade on a single draw.
26 hearts or spades, 52 cards, 26/52 or 0.5
e. The probability of drawing a face card given that you drew a heart.
13 hearts, 3 are face cards. 3/13 or 0.231
f. The probability of drawing a spade given that you drew a even number (2,4,6,8,10).
There are 20 even numbers in the deck and 5 are spades. 5/20 or 0.25
g. The probability of drawing a heart and a spade on consecutive draws.
13 of each. The first draw is out of 52 cards and the second will be out of 51. So we
have (13/52)*(13/51) = 0.063
h. The probability of drawing a non-face spade on the first draw and a face card on the
second, no replacement of the first draw.
There are 10 spades that are not face cards, so draw #1 is 10/52. The second asks us for
one of the 12 face cards out of the 51 remaining. So we have (10/52)*(12/51) = 0.045
i. The probability of drawing a 4 on the first draw and an adjacent number (a 3 or a 5) on
the second draw, no replacement of any draws.
Draw #1 is 4/52 as there are four 4’s in the deck. Of the remaining 51 cards for draw #2,
we are interested in 8 possible. So it’s (4/52)*(8/51) = 0.012
j. The probability of drawing a 4 on the first draw and an even number on the second
draw, replacing the first card you drew.
The new wrinkle here is that we are putting the original four back. This affects the
denominator of draw #2 (making it 52) as well as the numerator as we’re putting the
even-numbered 4 back in. So we have (4/52)*(20/52) = 0.029
k. The probability of being dealt four of a kind on a five card hand (this one’s tough).
After 30 minutes of wrangling, I can’t get this calculated without combination formulas
that we won’t be covering in this class. The actual number is 0.000240
l. The probability of drawing a royal flush of any suit with five consecutive draws of cards,
no replacement of any draws (so is this one).
The trick here is that I said “any suit”. The initial draw will determine which suit you’re
after. Since a royal flush is A,K,Q,J, 10 of the same suit, our first draw has 20
“successes”. After the first draw your suit is determined leaving one each of the
remaining values (K,Q,J,10 if you drew the A first). So we have
“It’s supposed to be hard. Otherwise, everyone would do it.
The hard is what makes it great.”
-Tom Hanks as Jimmy Dugan, A League of Their Own
Name:___________________________________
(20/52)*(4/51)*(3/50)*(2/49)*(1/48) = 0.0000015
“It’s supposed to be hard. Otherwise, everyone would do it.
The hard is what makes it great.”
-Tom Hanks as Jimmy Dugan, A League of Their Own
Homework #3 (Day)
Due: Mon, Jun 13, 11 am